AP EAMCET · PHYSICS · Center of Mass Momentum and Collision
A circular disc of radius \(R\) is removed from one end of a bigger circular disc of radius \(2 R\). The centre of mass of the new disc is at a distance \(\alpha R\) from the centre of the bigger disc. The value of \(\alpha\) is
- A \(\frac{1}{2}\)
- B \(\frac{1}{3}\)
- C \(\frac{1}{4}\)
- D \(\frac{1}{6}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{3}\)
Step-by-step Solution
Detailed explanation
Let, mass of entire disc \(=M\) Mass per unit area \(=\frac{M}{\pi(2 R)^2}=\frac{M}{4 \pi R^2}\) Mass of removed disc of radius \(R\), \( M_1=\frac{M}{4 \pi R^2} \times R^2=\frac{\mathrm{M}}{4} \) Mass of remaining disc, \( \Rightarrow \quad M_2=M-\frac{M}{4}=\frac{3 M}{4} \)…
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